tough hollow
I understand that i=3 goes to twenty and i=5 goes to 20 and we write them out and i=3,4,5,6...20 and i=5,6,7,8 and cancel out the common terms leaving us with 3^ and 4^2 which gives us 25, my question is this since i=5 in the second summation has a fraction 1/k, should i=5 give us 1/5, 1/6,i/7 up to 1/20? in this case how do you subtract the common terms from but sums if the sums from i=3 are 3,4,5...20 and the sums from i=5 are 1/5, 1/6, 1/7...1/20?
tough quiverBOT
placid quartz
I understand that i=3 goes to twenty and i=5 goes to 20 and we write them out an...
The answer to your question is in the expression in the middle
In short, you regroup the common terms and, for each i from 5 to 20, you count both the i^2 and the 1/i
tough hollow
I know the answer is in the middle (25), but how did we subtract i=3 which gave us numbers like 3^2,4^2...20^2 from k=5 which gave us fractions 1/5,1/6,1/7....1/20?
tough hollow
someone explained this to me as "there are no common terms, there was no subtraction between both summations, we just wrote the first 2 terms on the first sum to get to i=5
3²+4² = 25,so basically when the summation is mixed (one summation the sum of squares and the other a fraction) all we have to do is write the first terms from the starting point of the first summation (i=3 in this example) to the starting point of the second summation (i=5 in this example)"
is this correct?
placid quartz
someone explained this to me as "there are no common terms, there was no subtrac...
Okay, I didn't understand your question this way. Yeah basically that's what that person said
One one hand you have a set of 18(=20-3+1) numbers ordered from 3 to 20 (the i²) and on the other you have another set of 16(=20-5+1) numbers ordered from 5 to 20.
You want to regroup numbers from both sets that have numbers 5 to 20, so you have put apart number 3 and number 4 of the first set